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\newcommand{\CourseName}{高等代数一考试B}
\newcommand{\CourseClass}{2023级数学与应用数学、应用统计学}
\newcommand{\CourseDate}{2024年3月}

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\begin{document}

\maketitle

\thispagestyle{fancy} % 第一页也显示“第几页，共几页”的信息。

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%% 题目和解答从这里开始

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\begin{enumerate}

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%\newpage 
\item  %选择题第1题
在关于 $x$ 的多项式 $f(x)=\left|\begin{array}{cccc}2 & x & -5 & 3 \\ 1 & 2 & 3 & 4 \\ -1 & 0 & -2 & -3 \\ -1 & 7 & -2 & -2\end{array}\right|$ 中, 一次项的系数是什么？\dotfill (\qquad)
              
\begin{tasks}(4) %每行4个选项
\task [A.]  $-1$
\task [B.]  $1$
\task [C.]  $-2$
\task [D.]  $2$
\end{tasks}

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%%选择题第1题：解答
\ifnum\showsolution>0

{\color{red} 解答：A. 将行列式按第一行展开， 可得 
\begin{equation*}
f(x) = 2A_{11} + xA_{12} + (-5)A_{13} + 3A_{14}, 
\end{equation*}
其中 $A_{ij}$ 是第 $i$ 行、第 $j$ 列的元素 $a_{ij}$ 的代数余子式。
因为这个行列式的第2、3、4行都没有 $x$, 因此所求的 $x$ 的系数就是 $A_{12}$, 根据代数余子式的定义，计算可得
\begin{equation*}
A_{12} = (-1)^{1+2}\begin{vmatrix} 1 & 3 & 4 \\ -1 & -2 & -3 \\ -1 & -2 & -2 \end{vmatrix} = -1. 
\end{equation*}
}

\fi

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%\newpage 
\item  %选择题第2题
设$\boldsymbol{\eta}_{1}, \boldsymbol{\eta}_{2}$ 为线性方程组 $\boldsymbol{Ax}=\boldsymbol{\beta}$的解, $\boldsymbol{A}$ 的秩为$2$, $\boldsymbol{\eta}_{1}=(1,0,2)^{\mathrm{T}}, \boldsymbol{\eta}_{1}+\boldsymbol{\eta}_{2}=(1,-2,1)^{\mathrm{T}}$, 则对任意常数 $k$, 方程组 $\boldsymbol{A x}=\boldsymbol{\beta}$ 的通解是什么？ \dotfill (\qquad)


\begin{tasks}(2) %每行2个选项
\task [A.]  $(1,0,2)^{\mathrm{T}}+k(1,-2,1)^{\mathrm{T}}$
\task [B.]  $(0,-2,-1)^{\mathrm{T}}+k(1,2,3)^{\mathrm{T}}$
\task [C.]  $(2,0,4)^{\mathrm{T}}+k(1,-2,1)^{\mathrm{T}}$
\task [D.]  $(1,-2,1)^{\mathrm{T}}+k(2,0,4)^{\mathrm{T}}$
\end{tasks}
              
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%%选择题第2题：解答
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{\color{red} 解答：B. 
根据解向量的长度可知未知数的个数 $n=3$. 
根据 $\boldsymbol{\eta}_{1}, \boldsymbol{\eta}_{2}$ 为线性方程组 $\boldsymbol{Ax}=\boldsymbol{\beta}$的解可知 
$\boldsymbol{\eta}_{1}-\boldsymbol{\eta}_{2}= 2\boldsymbol{\eta}_{1}-( \boldsymbol{\eta_1} + \boldsymbol{\eta}_{2}) =(1,2,3)^{\mathrm{T}}$ 为齐次线性方程组 $\boldsymbol{Ax}=\boldsymbol{0}$的解。
因为 $n-r(\boldsymbol{A})=3-2=1$, 所以 $\boldsymbol{\eta}_{1}-\boldsymbol{\eta}_{2}$ 是 $\boldsymbol{Ax}=\boldsymbol{0}$ 的一个基础解系。
$\boldsymbol{\eta}_{2}=(0,-2,-1)^{\mathrm{T}}$ 是线性方程组 $\boldsymbol{Ax}=\boldsymbol{\beta}$的一个特解。

}

\fi

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%\newpage 
\item  %选择题第3题
当 $t$ 等于多少时, 向量组 $\boldsymbol{\alpha}_{1}=(2,1,0), \boldsymbol{\alpha}_{2}=(3,2,5), \boldsymbol{\alpha}_{3}=(5,4, t)$ 线性相关？ \dotfill (\qquad)

              
\begin{tasks}(4) %每行4个选项
\task [A.]  $10$
\task [B.]  $15$
\task [C.]  $-25$
\task [D.]  $25$
\end{tasks}

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%%选择题第3题：解答
\ifnum\showsolution>0

{\color{red} 解答：B. 
按定义，向量组 $\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 线性相关，
是指存在不全为零的实数 $x_1,x_2,x_3$ 使得 
$x_1\boldsymbol{\alpha}_{1} + x_2\boldsymbol{\alpha}_{2}+ x_3 \boldsymbol{\alpha}_{3} = \boldsymbol{0}$. 
这等价于齐次线性方程组
\begin{equation*}
(\boldsymbol{\alpha}_{1}^{\mathrm{T}}, \boldsymbol{\alpha}_{2}^{\mathrm{T}}, \boldsymbol{\alpha}_{3}^{\mathrm{T}})
\begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = \boldsymbol{0}
\end{equation*}
有非零解。这等价于系数矩阵的行列式的值等于零。因此有
\begin{equation*}
\begin{vmatrix} 2&3&5 \\ 1&2&4 \\ 0&5&t \end{vmatrix} = 0. 
\end{equation*}


}

\fi

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%\newpage 
\item  %选择题第4题
设矩阵 $\boldsymbol{A}$ 和 $\boldsymbol{B}$ 等价, $\boldsymbol{A}$ 有一个 $k$ 阶子式不等于零, 则关于 $\boldsymbol{B}$ 的秩正确的是哪个？ \dotfill (\qquad)

\begin{tasks}(4) %每行4个选项
\task [A.]  $r(\boldsymbol{B})\leqslant k$
\task [B.]  $r(\boldsymbol{B})\geqslant k$
\task [C.]  $r(\boldsymbol{B})=k$
\task [D.]  以上都不对
\end{tasks}

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%%选择题第4题：解答
\ifnum\showsolution>0

{\color{red} 解答：B. 
因为矩阵 $\boldsymbol{A}$ 和 $\boldsymbol{B}$ 等价，所以 $r(\boldsymbol{A}) = r(\boldsymbol{B})$. 
因为矩阵 $\boldsymbol{A}$ 有一个 $k$ 阶子式不等于零，所以$r(\boldsymbol{A}) \ge k$. 
所以 $r(\boldsymbol{B})\geqslant k$. 

}

\fi

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%\newpage 
\item  %选择题第5题
实二次型 $f\left(x_{1}, \cdots, x_{n}\right)$ 的矩阵满足下述哪个条件时，必是正定的？\dotfill (\qquad)

\begin{tasks}(2) %每行2个选项
\task [A.]  实对称且所有元素都为正数
\task [B.]  实对称且主对角线上元素都为正数
\task [C.]  实对称且行列式的值为正数
\task [D.]  实对称且正惯性指数等于$n$
\end{tasks}

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%%选择题第5题：解答
\ifnum\showsolution>0

{\color{red} 解答：D. 
实二次型 $f\left(x_{1}, \cdots, x_{n}\right)$ 的正惯性指数等于 $n$, 是指这个实二次型的规范形是
$y_1^2 + y_2^2 + \cdots + y_n^2$. 
根据规范形的定义，存在实数范围内的非退化的线性替换 $\boldsymbol{x}=C\boldsymbol{y}$ 使得 
$$f(\boldsymbol{x}) = f(C\boldsymbol{y}) = y_1^2 + y_2^2 + \cdots + y_n^2. $$ 
因此对任意实数向量 $\boldsymbol{x}$ 均有 $f(\boldsymbol{x})\ge 0$, 且仅当 $\boldsymbol{x}=\boldsymbol{0}$ 时才有 $f(\boldsymbol{x})= 0$. 按定义这个实二次型是正定的。

}

\fi

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%选择题结束

\item  %填空题第1题
排列$135 \cdots (2n-3)(2n-1)(2n)(2n-2)(2n-4) \cdots 642$的逆序数是$\underline{\qquad \qquad}$.

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%%填空题第1题：解答
\ifnum\showsolution>0

{\color{red} 解答：$n(n-1)$. 
考虑 $n=4$, 这时的排列为 $1,3,5,7,8,6,4,2$. 
从左到右写出所有的逆序，可得
\begin{equation*}
\begin{aligned}
32, \\ 
54, 52, \\
76, 74, 72, \\ 
86, 84, 82, \\
64, 62, \\
42. 
\end{aligned}
\end{equation*}

可见逆序的个数为 $1+2+3+3+2+1=12$. 
对一般情况，容易推知逆序的个数为 $$1+2+\cdots + (n-1) + (n-1) +\cdots +2+1=n(n-1). $$


}

\fi

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%\newpage 
\item  %填空题第2题
设 $\boldsymbol{A}$ 是 $m \times n$ 矩阵, 若 $r(\boldsymbol{A})=n$, 则齐次线性方程组 $\boldsymbol{Ax}=\boldsymbol{0} \underline{\qquad \qquad}$ 只有零解.(填写: 一定或不一定).

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%%填空题第2题：解答
\ifnum\showsolution>0

{\color{red} 解答：一定。
因为 $\boldsymbol{A}$ 是 $m \times n$ 矩阵，所以线性方程组 $\boldsymbol{Ax}=\boldsymbol{0}$ 的未知数个数等于 $n$. 
现在有 $n-R(A)=n-n=0$, 所以齐次线性方程组只有零解。

}

\fi

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%\newpage 
\item  %填空题第3题
设$3$阶行列式$|\boldsymbol{A}|$ 的值为 $\dfrac{1}{3}$, $\boldsymbol{A}^*$是 $\boldsymbol{A}$ 的伴随矩阵, 则$|(2\boldsymbol{A})^{-1}-3\boldsymbol{A}^*|=\underline{\qquad \qquad}$.

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%%填空题第3题：解答
\ifnum\showsolution>0

{\color{red} 解答：$-\frac{3}{8}$. 
根据 $\boldsymbol{A}^{-1} = \frac{1}{|\boldsymbol{A}|}\boldsymbol{A}^*$ 可得 
$\boldsymbol{A}^* = |\boldsymbol{A}| \boldsymbol{A}^{-1}$. 因此 
\begin{equation*}
\begin{aligned}
\left\lvert (2\boldsymbol{A})^{-1}-3\boldsymbol{A}^* \right\rvert 
&= \left\lvert \frac{1}{2}\boldsymbol{A}^{-1}-3|\boldsymbol{A}|\boldsymbol{A}^{-1} \right\rvert 
= \left\lvert \frac{1}{2}\boldsymbol{A}^{-1}-\boldsymbol{A}^{-1} \right\rvert \\ 
&= \left\lvert -\frac{1}{2}\boldsymbol{A}^{-1} \right\rvert 
= \left(-\frac{1}{2}\right)^3 \left\lvert\boldsymbol{A}^{-1} \right\rvert 
= -\frac{1}{8}\cdot 3 = -\frac{3}{8}. 
\end{aligned}
\end{equation*}

}

\fi

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%\newpage 
\item  %填空题第4题
初等矩阵的逆矩阵$\underline{\qquad \qquad}$是初等矩阵.(填写: 一定或不一定).

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%%填空题第4题：解答
\ifnum\showsolution>0

{\color{red} 解答：一定。
按定义，单位矩阵经过一次初等变换得到的矩阵称为初等矩阵。
单位矩阵经过一次``反方向''的初等变换也得到一个初等矩阵。
这两个初等矩阵互为逆阵。
这里``反方向''的初等变换是指：
\begin{enumerate}[label={(\arabic*)}]
\item  交换两行 $\longleftrightarrow$ 交换同样两行；
\item  将某一行乘以非零常数 $c$ $\longleftrightarrow$ 将这一行乘以非零常数 $\frac{1}{c}$; 
\item  将某一行乘以常数 $k$ 加到另一行 $\longleftrightarrow$ 将同样这一行乘以常数 $-k$ 加到同样另一行。 
\end{enumerate} 
}

\fi

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%\newpage 
\item  %填空题第5题
$n$ 阶实对称矩阵按合同分类, 共有 $\underline{\qquad \qquad}$ 类.

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%%填空题第5题：解答
\ifnum\showsolution>0

{\color{red} 解答：$\frac{(n+1)(n+2)}{2}$. 
$n$ 阶实对称矩阵的规范形为 $$\begin{pmatrix} \boldsymbol{E}_p&\boldsymbol{O}&\boldsymbol{O} \\ \boldsymbol{O}&-\boldsymbol{E}_q&\boldsymbol{O} \\ \boldsymbol{O}&\boldsymbol{O}&\boldsymbol{O}_{k} \end{pmatrix}, $$
其中 $p+q+k=n$. 因此不同的规范形的个数等于 $p+q+k= n$ 的在$[0,n]$ 范围内的整数解的个数。
\begin{enumerate}[label={(\arabic*)}]
\item  当 $k=0$ 时，$(p,q)=(n,0), (n-1,1), \cdots, (1,n), (0,n)$ 共有 $n+1$ 种情况；
\item  当 $k=1$ 时，$(p,q)= (n-1,0), (n-2,1), \cdots, (0,n-1)$ 共有 $n$ 种情况；$\cdots$
\item  当 $k=n$ 时，$(p,q)=(0,0)$ 共有1种情况。
\end{enumerate} 
因此不同的规范形的个数为 $ (n+1) + n + \cdots + 2+1=\frac{1}{2}(n+1)(n+2). $
}

\fi

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%填空题结束

\item %计算题第1题
设$\boldsymbol{A}$为$n$阶方阵, 如果$|\boldsymbol{A}|\neq 0$, 称$\boldsymbol{A}$为非退化的. 请叙述8条"矩阵$\boldsymbol{A}$非退化"的充分必要条件(要求: 每一条都不相同).

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%%计算题第1题：解答
\ifnum\showsolution>0

{\color{red}解答：
下列均为 $n$ 阶矩阵 $\boldsymbol{A}$ 非退化的充要条件: 
%(每条$\textcolor{red}{(+1)}$)
\ifnum\showsolution=2 \dotfill (\, 每条1分\, ) \fi

\begin{enumerate}[label={(\arabic*)}]

% \item  $|\boldsymbol{A}| \neq 0$;

\item  $\boldsymbol{A}_{n \times n}$可逆;

\item  $\boldsymbol{A}_{n \times n}$非奇异;  

\item  $\boldsymbol{A}_{n \times n}^{\rm{T}}$ 可逆;  

\item  $\boldsymbol{Ax}=\boldsymbol{0}$ 只有零解;  

\item  对任一$n$维向量 $\boldsymbol{\beta}$, $\boldsymbol{Ax}=\boldsymbol{\beta}$ 有唯一解;  

\item  对任一$n$维向量 $\boldsymbol{\beta}$, $\boldsymbol{Ax}=\boldsymbol{\beta}$ 都有解;  

\item  $\boldsymbol{A}$的秩等于$n$;  

\item  $\boldsymbol{A}$的行秩等于$n$ (或$\boldsymbol{A}$ 行满秩); 

\item  $\boldsymbol{A}$的列秩等于$n$ (或$\boldsymbol{A}$ 列满秩); 

\item  $\boldsymbol{A}$ 的行向量组线性无关; 

\item  $\boldsymbol{A}$ 的列向量组线性无关; 

\item  任一 $n$ 维向量可由 $\boldsymbol{A}$ 的行向量组线性表出; 

\item  任一 $n$ 维向量可由 $\boldsymbol{A}$ 的列向量组线性表出; 

\item  $n$ 维单位行向量组$\boldsymbol{\varepsilon}_1, \boldsymbol{\varepsilon}_2, \cdots, \boldsymbol{\varepsilon}_n$可由 $\boldsymbol{A}$ 的行向量组线性表出; 

\item  $n$ 维单位列向量组$\boldsymbol{e}_1, \boldsymbol{e}_2, \cdots, \boldsymbol{e}_n$可由 $\boldsymbol{A}$ 的列向量组线性表出; 

\item  $n$ 维单位行向量组$\boldsymbol{\varepsilon}_1, \boldsymbol{\varepsilon}_2, \cdots, \boldsymbol{\varepsilon}_n$与$\boldsymbol{A}$ 的行向量组等价; 

\item  $n$ 维单位列向量组$\boldsymbol{e}_1, \boldsymbol{e}_2, \cdots, \boldsymbol{e}_n$与$\boldsymbol{A}$ 的列向量组等价; 

\item  $\boldsymbol{A}$ 的(等价)标准形为单位矩阵(或 $\boldsymbol{A}$ 与单位矩阵等价, 或$\boldsymbol{A}$ 可以经过一系列初等变换化为单位矩阵); 

\item  $\boldsymbol{A}$ 能表示成一些(或“有限个”)初等矩阵的乘积; 

\item  $\boldsymbol{A}$ 的伴随矩阵$\boldsymbol{A}^*$可逆(再根据如上条件又可写出一系列充要条件, 比如$(\boldsymbol{A}^*)^{\rm{T}}$可逆, $\boldsymbol{A}^*$ 的伴随矩阵$\boldsymbol{A}^{**}$可逆, 如此进行下去, 可写出无穷多充要条件).  

\end{enumerate} 

}

\vspace{0.2cm}

\fi

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%\newpage 
\item %计算题第2题
求$n(n>1)$ 阶方阵$\boldsymbol{A}$ 的逆, 其中
\[\boldsymbol{A}=\left(\begin{array}{cccccc}
1 & -1 & 1 & \cdots  & (-1)^{n-2} & (-1)^{n-1} \\
0 & 1 & -1 & \cdots  & (-1)^{n-3} & (-1)^{n-2} \\
0 & 0 &  1 & \cdots  & (-1)^{n-4} & (-1)^{n-3} \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots  & 1 & -1  \\
0 & 0 & 0 & \cdots  & 0 & 1
\end{array}\right) .\]

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%%计算题第2题：解答
\ifnum\showsolution>0

{\color{red}解答：初等变换法: 构造矩阵 $\left(\boldsymbol{A}, \boldsymbol{E}\right)\xrightarrow[i=1, 2, \cdots, n-1]{r_i+r_{i+1}}\left(\boldsymbol{E}, \boldsymbol{A}^{-1}\right)$, 
%$\textcolor{red}{(+4)}$
\fourpoints

即得
\[\boldsymbol{A}^{-1}=\left(\begin{array}{cccccc}
1 & 1 & 0 & \cdots & 0 & 0 \\
0 & 1 & 1 & \cdots & 0 & 0 \\
0 & 0 & 1 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & & \vdots & \vdots \\
0 & 0 & 0 & \cdots & 1 & 1 \\
0 & 0 & 0 & \cdots & 0 & 1
\end{array}\right). \]
%\textcolor{red}{(+4)}
\fourpoints

}

\vspace{0.2cm}

\fi

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%\newpage 
\item %计算题第3题
讨论参数$k$为何值时, 下列方程组无解, 有唯一解, 有无穷多解? 在有唯一解时, 求出此解; 在有无穷多解时, 利用其导出组的基础解系写出全部的解.
\[\left\{\begin{array}{l}
k x_{1}+x_{2}+x_{3}=-2, \\
x_{1}+k x_{2}+x_{3}=-2, \\
x_{1}+x_{2}+k x_{3}=-2 .
\end{array}\right.\]

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%%计算题第3题：解答
\ifnum\showsolution>0

{\color{red}解答：对增广矩阵作初等行变换:
\[(\boldsymbol{A},\boldsymbol{\beta})=\left(\begin{array}{cccc}
k & 1 & 1 & -2  \\
1 & k & 1 & -2 \\
1 & 1 & k & -2
\end{array}\right)  \rightarrow
\left(\begin{array}{cccc}
1 & k & 1 & -2  \\
k & 1 & 1 & -2 \\
1 & 1 & k & -2
\end{array}\right) \rightarrow 
\left(\begin{array}{cccc}
1 & k & 1 & -2  \\
0 & 1-k^2 & 1-k & -2+2k \\
0 & 1-k & k-1 & 0
\end{array}\right)  \rightarrow\]
\[
\left(\begin{array}{cccc}
1 & k & 1 & -2  \\
0 & 0 & -k^2-k+2 & 2k-2 \\
0 & 1-k & k-1 & 0
\end{array}\right)  \rightarrow
\left(\begin{array}{cccc}
1 & k & 1 & -2  \\
0 & 1-k & k-1 & 0 \\
0 & 0 & -(k-1)(k+2) & 2(k-1)              
\end{array}\right). \]
%\textcolor{red}{(+2)}
\twopoints
\begin{enumerate}[label={(\arabic*)}]
\item %(1) 
当 $k=-2$ 时, 方程组无解; %$\textcolor{red}{(+2)}$ 
\twopoints

\item %(2) 
当 $k=1$ 时, 方程组的全部的解为 $(-2,0,0)^{\rm{T}}+c_{1}(-1,1,0)^{\rm{T}}+$ $c_{2}(-1,0,1)^{\rm{T}}$, 其中 $c_{1}, c_{2}$ 为任意常数; 
%$\textcolor{red}{(+2)}$
\twopoints

\item %(3) 
当 $k \neq-2,1$ 时, 方程组有唯一解 $\left(-\dfrac{2}{k+2},-\dfrac{2}{k+2},-\dfrac{2}{k+2}\right)^{\rm{T}}$. 
%$\textcolor{red}{(+2)}$
\twopoints

\end{enumerate}

}

\vspace{0.2cm}

\fi

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%\newpage 
\item %计算题第4题
求 $a, b, c$ 的取值范围, 使得下列实二次型$f\left(x_{1}, x_{2}, x_{3}\right)=a x_{1}^{2}+b x_{2}^{2}+a x_{3}^{2}+2 c x_{1} x_{3}$是正定的.

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%%计算题第4题：解答
\ifnum\showsolution>0

{\color{red}解答：
二次型$f$的矩阵为 $$\boldsymbol{A}=\left(\begin{array}{ccc}
a & 0 & c \\
0 & b & 0 \\
c & 0 & a
\end{array}\right).$$
 
因为 $f$ 是正定二次型，所以 $\boldsymbol{A}$ 是正定矩阵，所以 $\boldsymbol{A}$ 的三个顺序主子式全都大于零，即
\begin{equation*}
a>0,  \\ 
%\textcolor{red}{(+2)}
a b>0,  \\
%\textcolor{red}{(+2)}
(a^{2}-c^{2}) b>0. 
%\textcolor{red}{(+2)} 
\end{equation*}
\ifnum\showsolution=2 \dotfill (\, 每个不等式 2 分，共6分 \, ) \fi

解得 $a, b, c$ 需满足的条件为 $a>0, b>0$ 且 $a>|c|$. 
%$\textcolor{red}{(+2)}$
\twopoints

}

\vspace{0.2cm}

\fi

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%\newpage 
\item %计算题第5题
化实二次型 $f\left(x_{1}, x_{2}, x_{3}\right)=x_{1} x_{2}+x_{1} x_{3}+x_{2} x_{3}$ 为规范形, 并写出所作的非退化线性替换.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%计算题第5题：解答
\ifnum\showsolution>0

{\color{red}解答：
记$\boldsymbol{X}=\left(x_{1}, x_{2}, x_{3}\right)^{\mathrm{T}}$为$3$维实向量, 则 $f(\boldsymbol{X})=\boldsymbol{X}^{\mathrm{T}} \boldsymbol{A} \boldsymbol{X}$, 其中 
$$\boldsymbol{A}=\left(\begin{array}{ccc}0 & \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & 0 & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} & 0\end{array}\right).$$

构造分块矩阵，对矩阵 $\boldsymbol{A}$ 进行初等行变换和初等列变换，并将初等列变换记录在下方的单位矩阵里，
\begin{equation*}
\begin{aligned}
\left(\begin{array}{c}\boldsymbol{A} \\ \boldsymbol{E}\end{array}\right)
&=\left(\begin{array}{ccc}
0 & \frac{1}{2} & \frac{1}{2}  \\ \frac{1}{2} & 0 & \frac{1}{2}  \\ \frac{1}{2} & \frac{1}{2} & 0 \\  1&0&0 \\  0&1&0 \\ 0&0&1
\end{array}\right)
\xrightarrow[r_1+r_2]{c_1+c_2} 
\left(\begin{array}{ccc}
1 & \frac{1}{2} & 1  \\ \frac{1}{2} & 0 & \frac{1}{2}  \\ 1 & \frac{1}{2} & 0 \\  1 & 0 & 0 \\  1 & 1 & 0 \\ 0 & 0 & 1
\end{array}\right)
\xrightarrow[r_2-\frac{1}{2}r_1]{c_2-\frac{1}{2}c_1} 
\left(\begin{array}{ccc}
1 & 0 & 1  \\ 0 & -\frac{1}{4} & 0  \\ 1 & 0 & 0 \\  1 & -\frac{1}{2} & 0 \\  1 & \frac{1}{2} & 0 \\ 0 & 0 & 1
\end{array}\right) \\ 
& \xrightarrow[r_3-r_1]{c_3-c_1} 
\left(\begin{array}{ccc}
1 & 0 & 0  \\ 0 & -\frac{1}{4} & 0  \\ 0 & 0 & -1 \\  1 & -\frac{1}{2} & -1 \\  1 & \frac{1}{2} & -1 \\ 0 & 0 & 1
\end{array}\right)
\xrightarrow[r_2 \times 2]{c_2 \times 2} 
\left(\begin{array}{ccc}
1 & 0 & 0  \\ 0 & -1 & 0  \\ 0 & 0 & -1 \\  1 & -1 & -1 \\  1 & 1 & -1 \\ 0 & 0 & 1
\end{array}\right)
=\left(\begin{array}{c}\boldsymbol{B} \\ \boldsymbol{C}\end{array}\right). 
\end{aligned}
\end{equation*}
%$\textcolor{red}{(+4)}$
\fourpoints

记
\begin{equation*}
\boldsymbol{B}=\left(\begin{array}{ccc} 1 & 0 & 0  \\ 0 & -1 & 0  \\ 0 & 0 & -1 \end{array}\right), 
\boldsymbol{C}=\left(\begin{array}{ccc} 1 & -1 & -1 \\  1 & 1 & -1 \\ 0 & 0 & 1 \end{array}\right), 
\boldsymbol{Y}=\begin{pmatrix} y_{1} \\ y_{2} \\ y_{3} \\ \end{pmatrix}, 
\end{equation*}
则 $\boldsymbol{B}=\boldsymbol{C}^{\mathrm{T}} \boldsymbol{A} \boldsymbol{C}$.
故二次型 $f(x_{1}, x_{2}, x_{3})$ 经非退化线性替换 $\boldsymbol{X}=\boldsymbol{CY}$ 可化为规范形
$$g(y_{1}, y_{2}, y_{3}) = y_{1}^{2} - y_{2}^{2} - y_{3}^{2}.$$ 
%$\textcolor{red}{(+4)}$
\fourpoints

}

\vspace{0.2cm}

\fi

%%计算题结束
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\item %证明题第1题
设$\boldsymbol{A}$ 是 $n \times n$ 矩阵 $(n \geqslant 2)$, $\boldsymbol{A}^{*}$是其伴随矩阵. 
\begin{enumerate}[label={(\arabic*)}]
\item % (1) 
证明: $\left|\boldsymbol{A}^{*}\right|=|\boldsymbol{A}|^{n-1}$;
 
\item % (2) 
若$\boldsymbol{A}$不可逆, 证明: $r\left(\boldsymbol{A})+r(\boldsymbol{A}^{*}\right) \leqslant n.$
 
\item % (3) 
证明: $r\left(\boldsymbol{A}^{*}\right)=\left\{\begin{array}{ll}                 
 n, & r(\boldsymbol{A})=n, \\                 
 1, & r(\boldsymbol{A})=n-1, \\                 
 0, & r(\boldsymbol{A})<n-1 .                 
 \end{array}\right.$    

\end{enumerate}
   
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%%证明第1题：解答
\ifnum\showsolution>0

{\color{red}证明：
\begin{enumerate}[label={(\arabic*)}]

\item %(1) 
由$\boldsymbol{A} \boldsymbol{A}^{*}=|\boldsymbol{A}| \boldsymbol{E}_n$, 两边同时取行列式，根据行列式乘积公式，可得
 \[|\boldsymbol{A}|\left|\boldsymbol{A}^{*}\right| 
 = \left|\boldsymbol{A} \boldsymbol{A}^{*}\right|
 = \left| |\boldsymbol{A}| \boldsymbol{E}_n \right| 
 = |\boldsymbol{A}|^{n},\]
 
 \begin{enumerate}[label={(\roman*)}]
 \item %(i)
 若 $|\boldsymbol{A}| \neq 0$, 则 $\left|\boldsymbol{A}^{*}\right|=|\boldsymbol{A}|^{n-1}$. 
 %$\textcolor{red}{(+3)}$
\threepoints

 \item %(ii) 
 若 $|\boldsymbol{A}|=0$, 则 $\boldsymbol{A} \boldsymbol{A}^{*}=|\boldsymbol{A}| \boldsymbol{E}=\boldsymbol{O}$. 下面只需证 $\left|\boldsymbol{A}^{*}\right|=|\boldsymbol{A}|^{n-1}=0$.

 若不然, 假设$\left|\boldsymbol{A}^{*}\right| \neq 0$, 则 $\boldsymbol{A}^{*}$ 可逆, $\left(\boldsymbol{A}^{*}\right)^{-1}$ 存在, 则
 \[\boldsymbol{A} \boldsymbol{A}^{*}=|\boldsymbol{A}| \boldsymbol{E}=\boldsymbol{O} \Rightarrow \boldsymbol{A} \boldsymbol{A}^{*}\left(\boldsymbol{A}^{*}\right)^{-1}=\boldsymbol{O} \Rightarrow \boldsymbol{A}=\boldsymbol{O} \Rightarrow \boldsymbol{A}^{*}=\boldsymbol{O}.\]
 此与假设$\left|\boldsymbol{A}^{*}\right| \neq 0$矛盾. 故假设不成立, 因而$\left|\boldsymbol{A}^{*}\right|=0$. 
 %$\textcolor{red}{(+3)}$
\threepoints

\end{enumerate}

\item % (2) 
若$\boldsymbol{A}$不可逆, 则$|\boldsymbol{A}|=0$, $\boldsymbol{A} \boldsymbol{A}^{*}=|\boldsymbol{A}| \boldsymbol{E}=\boldsymbol{O}$. 由$\boldsymbol{A A}^*=\boldsymbol{O}$, 可知$\boldsymbol{A}^*$的每个列向量都是齐次线性方程组 $\boldsymbol{A x}=\mathbf{0}$ 的解. 
%$\textcolor{red}{(+3)}$
\threepoints

又因为方程组 $\boldsymbol{A x}=\mathbf{0}$ 的基础解系中的向量个数为 $n-r(\boldsymbol{A})$, 则 $\boldsymbol{A}^*$ 的列向量组的秩小于等于 $n-r(\boldsymbol{A})$, 即得 $r\left(\boldsymbol{A}^{*}\right) \leqslant n-r(\boldsymbol{A})$. 故$r(\boldsymbol{A})+r\left(\boldsymbol{A}^{*}\right) \leqslant n$. 
%$\textcolor{red}{(+3)}$
\threepoints

\item %(3) 
分三种情形证明: 
\begin{enumerate}[label={(\roman*)}]
\item %(i) 
当 $r(\boldsymbol{A})=n$时, $|\boldsymbol{A}| \neq 0, \left|\boldsymbol{A}^{*}\right|=|\boldsymbol{A}|^{n-1} \neq 0$. 故$r\left(\boldsymbol{A}^{*}\right)=n$. 
%$\textcolor{red}{(+2)}$
\twopoints

\item %(ii) 
当$r(\boldsymbol{A})=n-1$ 时, $\boldsymbol{A}$ 至少有一个 $n-1$ 阶子式不等于零, 即 $\boldsymbol{A}^{*} \neq \boldsymbol{O}$, $r\left(\boldsymbol{A}^{*}\right) \geqslant 1$. 

又因$r(\boldsymbol{A})=n-1, |\boldsymbol{A}|=0$, 故$\boldsymbol{A} \boldsymbol{A}^{*}=|\boldsymbol{A}| \boldsymbol{E}=\boldsymbol{O}$. 由(2)知, $r(\boldsymbol{A})+r\left(\boldsymbol{A}^{*}\right) \leqslant n$, 故$r\left(\boldsymbol{A}^{*}\right) \leqslant n-(n-1)=1$, 所以$r\left(\boldsymbol{A}^{*}\right)=1$. 
%$\textcolor{red}{(+4)}$
\fourpoints

\item %(iii) 
当$r(\boldsymbol{A})<n-1$ 时, $\boldsymbol{A}$ 的任意$n-1$阶子式都为零. 所以$\boldsymbol{A}^{*}=\boldsymbol{O}$, 因而$r\left(\boldsymbol{A}^{*}\right)=0$. 
%$\textcolor{red}{(+2)}$

\twopoints

\end{enumerate}

\end{enumerate}

}

\vspace{0.2cm}

\fi

%%证明题结束
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\end{enumerate}

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%%试卷结束
\end{document}





